Se F(x) = x2-5x+6, quanto vale F(3/4) e F(5/7) ?
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F(x) = 2x-5x + 6
F(x) = -3x + 6
f(3/4) = -3*3/4 + 6
F(3/4)= -9/4+6
F(3/4)= -9+24/4
f(3/4)= 15/4
F(3/4)= 3,75
F(5/7) = -3*5/7+6
F(5/7) = -15/7+6
F(5/7) = -15+42/6
F(5/7) = 27/6
F(5/7) = 4,5
F(x) = -3x + 6
f(3/4) = -3*3/4 + 6
F(3/4)= -9/4+6
F(3/4)= -9+24/4
f(3/4)= 15/4
F(3/4)= 3,75
F(5/7) = -3*5/7+6
F(5/7) = -15/7+6
F(5/7) = -15+42/6
F(5/7) = 27/6
F(5/7) = 4,5
amanda691:
obgriado :))
por nada
é 2x ou x²?
vish, eu vi 2X agora se for x² vc está certa.
x ao quadrado
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f(3/4)= (3/4)²-5*(3/4)+6
f(3/4)= (3/4)*(3/4)-(15/4)+6
f(3/4)= (9/16)-(15/4)+6
MMC de 16 e 4 = 16
f(3/4)= (9-60+96)/16
f(3/4)= 45/16
f(5/7)= (5/7)²-5(5/7)+6
f(5/7)= (5/7)*(5/7)-(25/7)+6
f(5/7)= (25/49)-(25/7)+6
MMC de 49 e 7 = 49
f(5/7)= (25-175+294)/49
f(5/7)= 144/49
f(3/4)+f(5/7)=
46/16 + 144/49
MMC de 16 e 49 = 784
(2254+2304)/784
4558/784
f(3/4)= (3/4)*(3/4)-(15/4)+6
f(3/4)= (9/16)-(15/4)+6
MMC de 16 e 4 = 16
f(3/4)= (9-60+96)/16
f(3/4)= 45/16
f(5/7)= (5/7)²-5(5/7)+6
f(5/7)= (5/7)*(5/7)-(25/7)+6
f(5/7)= (25/49)-(25/7)+6
MMC de 49 e 7 = 49
f(5/7)= (25-175+294)/49
f(5/7)= 144/49
f(3/4)+f(5/7)=
46/16 + 144/49
MMC de 16 e 49 = 784
(2254+2304)/784
4558/784
é x ao quadrado rs, obg
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